Is there some suitable generalization to the notion of Baire property for topological spaces of arbitrary cardinalities which satisfies the following condition:

The meager sets are sets which are union of $\lambda$ nowhere dense sets where $\lambda < \kappa$ and $\kappa$ is the cardinality of the space.

If we consider a model of ZFC $\mathfrak{M}$ in which every (suitably definable) set of reals have the property of Baire, then every (suitably definable) set of a suitably defined toplogical space have the "generalized" Baire property.

The usual Baire property involves {\it countable} union of nowhere dens sets. Maybe there is a generalization to $\lambda$ union of nowhere dense sets, for $\lambda < \kappa$ where $\kappa$ is the cardinality of the space.
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user38200Dec 10 '13 at 19:19

I assume that we define the topological space in such a way that if it has cardinality $\kappa > 2^{\aleph_0}$ in $\mathfrak{M}$ and if it has cardinality $2^{\aleph_0}$ in the real world, the topology must coincide with the real topology.
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user38200Dec 10 '13 at 20:17

You may already know this, but the most well-known generalisation of this is probably to Polish spaces, i.e. separable completely metrisable spaces. I don’t know if “every set of reals is Baire” implies “every subset of a Polish space is Baire”, but the standard models of the former are also models of the latter, if I remember right. However, this is a bit short of a full answer, since Polish spaces cannot have cardinality larger than the continuum.
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Peter LeFanu LumsdaineDec 10 '13 at 20:47

@EmilJeřábek: Yes that is exactly what I mean. For the models of ZFC, we can take the full Solovay model or a model of projective determinacy. Sorry I made some changes.
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user38200Dec 10 '13 at 21:09

(I’ve cleaned up my comments as they are no longer relevant.)
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Emil JeřábekDec 10 '13 at 23:38

1 Answer
1

I am not sure if this is the kind of answer you were looking for, but since no one has given an answer yet, I think it is a good idea to say what little I know about larger cardinal analogues of the Baire category theorem.

The only spaces that I can currently think of where one would want to consider larger cardinal generalizations of Baire spaces to larger cardinals are the $P_{\kappa}$-spaces, so I will say a few things about how Baire spaces relate to $P_{\kappa}$-spaces.
In order to give motivation for this answer, I will first have to say a few things about $P_{\kappa}$-spaces in general before I talk about the Baire property.

$\textbf{Preliminaries concerning $P_{\kappa}$-spaces}$

Many of the basic results from general topology hold when one replaces each instance of "finite" with an instance of "less than $\kappa$" for some regular cardinal $\kappa$. In this context, instead of dealing with ordinary topological spaces where the intersection of finitely many open sets is open, one deals with spaces where the intersection of less than $\kappa$ many open sets is open.

If $\lambda$ is a cardinal, then a $P_{\lambda}$-space is a completely regular space where the intersection of less than $\lambda$ many open sets is open. If $\lambda$ is a singular cardinal, then every $P_{\lambda}$ space is automatically a $P_{\lambda^{+}}$-space. Therefore, without loss of generality, we may restrict our attention to $P_{\lambda}$-spaces where $\lambda$ is a regular cardinal.

A space $X$ is said to be $\lambda$-compact if the intersection of less than $\lambda$ many open sets is open. If $\lambda$ is a regular cardinal, then the product of finitely many $\lambda$-compact $P_{\lambda}$-spaces is $\lambda$-compact.

One can also generalize the notion of a metric space and a uniform space to $P_{\kappa}$-spaces. We therefore define a $P_{\kappa}$-uniform space to be a uniform space where the intersection of less than $\kappa$ many entourages is an entourage. If $\kappa$ is uncountable, then it is easy to show that every $P_{\kappa}$-uniform space is generated by equivalence relations. It is well known that a uniform space is induced by a metric if and only if it is generated by a countable sequence of entourages. The notion of a metric space is analogous to the notion of a uniform space generated by linearly ordered descending sequence of entourages of length $\kappa$.

$\textbf{$P_{\kappa}$-spaces and Baire spaces}$

If $\lambda$ is an infinite cardinal, then a $\lambda$-Baire space is a a topological space $X$ such that the intersection of less than $\lambda$ many dense open sets is dense. With this definition, every topological space is an $\aleph_{0}$-Baire space, and an $\aleph_{1}$-Baire space is simple a Baire space.

Fortunately, the Baire category theorem does hold for some $P_{\kappa}$-spaces where you might expect there to be some form of the Baire category theorem. Let $\kappa$ be a regular cardinal. Suppose that $I$ is a set and $X_{i}$ is a space for $i\in I$. Let
$\prod_{i\in I}^{\kappa}X_{i}$ be the topology with underlying set $\prod_{i\in I}X_{i}$ generated by the basis consisting of products $\prod_{i\in I}U_{i}$ where each $U_{i}$ is open in $X_{i}$ and where $|\{i\in I|U_{i}\neq X_{i}\}|<\kappa$. If each $X_{i}$ is a $P_{\kappa}$-space, then $\prod_{i\in I}^{\kappa}X_{i}$ is also a $P_{\kappa}$-space.

$\mathbf{Proposition}$ If $I$ is a set, $\kappa$ is a regular
cardinal, and $X_{i}$ is a discrete space for each $i\in I$, then
$\prod_{i\in I}^{\kappa}X_{i}$ satisfies the $\kappa^{+}$-Baire
property.

$\mathbf{Proof}$ The proof of this result is analogous to the proof of the ordinary Baire category theorem. Let $U_{\alpha}\subseteq\prod_{i\in I}^{\kappa}X_{i}$ be a dense open set for each $\alpha<\kappa$ and let $U\subseteq\prod_{i\in I}^{\kappa}X_{i}$ be a nonempty open set. We shall using transfinite induction construct sets $I_{\alpha}\subseteq I$ with $|I_{\alpha}|<\kappa$ and functions $f_{\alpha}\in\prod_{i\in I_{\alpha}}X_{i}$ such that

if $\alpha<\beta$, then $I_{\alpha}\subseteq I_{\beta}$ and $f_{\alpha}=f_{\beta}|_{I_{\alpha}}$

If $f\in\prod_{i\in I}^{\kappa}X_{i}$, and $f|_{I_{\alpha}}=f_{\alpha}$, then
$f\in U\cap\bigcap_{\beta<\alpha}U_{\alpha}$.

Zero step- Since $U$ is a non-empty open set, there is some $I_{0}\subseteq I$ and some $f_{0}\in\prod_{i\in I_{0}}X_{i}$ where if $f\in\prod_{i\in I}X_{i}$ and $f|_{I_{0}}=f$.

Limit ordinal step- If $\lambda$ is a limit ordinal with $\lambda<\kappa$, then let $I_{\lambda}=\bigcup_{\alpha<\lambda}I_{\alpha}$ and let
$f_{\lambda}=\bigcup_{\alpha<\lambda}f_{\alpha}$.

Successor ordinal step- Suppose that $I_{\alpha},f_{\alpha}$ have been defined already and $|I_{\alpha}|<\kappa$. Then $\{f\in\prod_{i\in I}X_{i}:f|_{I_{\alpha}}=f_{\alpha}\}$ is a non-empty open set, so $U_{\alpha}\cap\{f\in\prod_{i\in I}X_{i}:f|_{I_{\alpha}}=f_{\alpha}\}$ is a non-empty open set. Therefore, there is some set $I_{\alpha+1}$ with $I_{\alpha}\subseteq I_{\alpha+1}$ and some $f_{\alpha+1}\in\prod_{i\in I_{\alpha+1}}X_{i}$ where if $f\in\prod_{i\in I}X_{i}$ and $f|_{I_{\alpha+1}}=f_{\alpha+1}$, then $f\in U_{\alpha}$.

Let $J=\bigcup_{\alpha<\kappa}I_{\alpha}$ and let $f\in\prod_{i\in I}X_{i}$ be a function with $f|_{J}=\bigcup_{\alpha<\kappa}f_{\alpha}$. Then $f\in U\cap\bigcap_{\alpha<\kappa}U_{\alpha}$. Therefore, since $U\cap\bigcap_{\alpha<\kappa}U_{\alpha}$ is non-empty, the set $\bigcap_{\alpha<\kappa}U_{\alpha}$ is dense. $\mathbf{QED}$

Interestingly, for $P_{\lambda}$-spaces, the property of being a $\lambda$-Baire space is hereditary under taking dense subspaces.

$\mathbf{Proposition}$ Let $\lambda$ be a cardinal and suppose that
$Y$ is a dense subspace of a space $X$.

If $Y$ is a $\lambda$-Baire space, then $X$ is also a $\lambda$-Baire space.

If $X$ is a $\lambda$-Baire $P_{\lambda}$-space, then $Y$ is also a $\lambda$-Baire space.

One can also construct $P_{\kappa}$-spaces with strong Baire properties using ultraproducts. Recall that an ultrafilter $U$ is $\lambda$-regular if there is some $E\subseteq U$ with $|E|=\lambda$ but where $\bigcap D=\emptyset$ for each infinite $D\subseteq E$. Furthermore, the countably incomplete $\lambda$-good ultrafilters are precisely the ultrafilters where the ultraproducts are always $\lambda$-saturated.

$\mathbf{Proposition}$ (Bankston)

The ultraproduct of regular spaces by a $\lambda$-regular ultrafilter is a $P_{\lambda^{+}}$-space.

Furthermore, the ultraproduct of topological spaces by a countably incomplete $\lambda$-good ultrafilter is $\lambda$-Baire.

$\textbf{Ideas of proof}$ A proof of these facts is given in the paper Topological reduced products via good ultrafilters by Paul Bankston. A proof of 1 uses basic facts about regular ultrafilters. A proof of 2 uses the fact that the ultraproduct of models by a $\lambda$-good ultrafilter is saturated.

$\textbf{The failure of the Baire property}$

One might expect for $\kappa$-compact $P_{\kappa}$-spaces to be Baire spaces, or complete uniform spaces generated by a linearly ordered set of entourages (i.e. generalized metric spaces) to be Baire spaces. Unfortunately, this is not the case. In fact, for most $P_{\kappa}$-spaces it is easy to find a closed subspace that is not even a Baire space. If $X$ is a $P_{\kappa}$-space, and there is a sequence of closed subsets $C_{n}$ where $C_{n}$ is nowhere dense in $C_{n+1}$ for all $n$, then $\bigcup_{n}C_{n}$ is a closed subspace of $X$ and each $C_{n}$ is nowhere dense in $\bigcup_{n}C_{n}$, so $\bigcup_{n}C_{n}$ is not a Baire space.

Let $X_{n}$ be a $P_{\kappa}$-space for all natural numbers $n>0$ and $x_{n}\in X_{n}$ be a non-isolated point for all $n$. Give $X=\prod_{n\in\mathbb{N}}X_{n}$ the box topology. Then $\prod_{n\in\mathbb{N}}X_{n}$ becomes a $P_{\kappa}$-space with this topology. For natural numbers $N$, let $C_{N}\subseteq\prod_{n\in\mathbb{N}}X_{n}$ consist of all sequences $(y_{n})_{n}$ such that $y_{n}=x_{n}$ whenever $n>N$. Then $C_{N}\simeq X_{1}\times...\times X_{N}$. Let $C=\bigcup_{N}C_{N}$. Then $C$ is not a Baire space. On the other hand, if each $X_{n}$ is $\kappa$-compact, then each finite product $X_{1}\times...\times X_{n}\simeq C_{n}$ is $\kappa$-compact, so $C$ is $\kappa$-compact. Furthermore, if each $X_{n}$ can be given a complete uniformity generated by a linearly ordered set of cofinality $\kappa$, then $C$ can also be given a
complete uniformity generated by a linearly ordered set of cofinality $\kappa$ (i.e. $C$ is like a complete metric space), but $C$ is still not a Baire space.

$\textbf{An application of generalized Baire spaces}$

In this section, I will give motivation for the Baire property for $P_{\kappa}$-spaces in terms of point-free topology and Boolean algebras. The notions in this section come from my own personal research.

Recall that a frame is a complete lattice that satisfies the distributive law $x\wedge\bigvee_{i\in I}y_{i}=\bigvee_{i\in I}(x\wedge y_{i})$. Frames are the main objects of study in point-free topology since if $(X,\mathcal{T})$ is a topological space, then $\mathcal{T}$ is a frame.

If $L,M$ are frames, then a frame homomorphism from $L$ to $M$ is a function $f:L\rightarrow M$ such that $f(0)=0,f(1)=1,f(\bigvee R)=\bigvee f[R],f(x\wedge y)=f(x)\wedge f(y)$ for each $R\subseteq L$ and $x,y\in L$. Frame homomorphisms from $L$ to $M$ correspond roughly to continuous functions from $M$ to $L$.

A frame $L$ shall be called $\kappa$-distributive if whenever $|I|<\kappa$ and $R_{i}\subseteq L$ for $i\in I$, then $\bigwedge_{i\in I}\bigvee R_{i}=\bigvee\{\bigwedge_{i\in I}x_{i}|x_{i}\in R_{i}\,\textrm{for}\,i\in I\}.$ It can be shown that a $T_{1}$-space is $\kappa$-distributive if and only if the intersection of less than $\kappa$ many open sets is open. Therefore, the notion of $\kappa$-distributivity is a way to generalize the notion of a $P_{\kappa}$-space to point-free topology. It turns out that the surjective image of a $\kappa$-distributive frame under a frame homomorphism is not necessarily $\kappa$-distributive. On the other hand, one may characterize the completely regular $\kappa$-distributive frames $L$ such that the image of $L$ under a surjective frame homomorphism is $\kappa$-distributive, and this characterization requires the Baire property. We shall state the results here for topological spaces.

$\mathbf{Proposition}$ (Sikorski) Let $X$ be a $P_{\kappa}$-space. Then the
regular open algebra $\mathrm{Ro}(X)$ is $\kappa$-distributive if and
only if $X$ is a $\kappa$-Baire space.

$\mathbf{Theorem}.$ Let $\kappa$ be a regular cardinal. Let
$(X,\mathcal{T})$ be a $P_{\kappa}$-space. Then the following are
equivalent.

Every closed subspace of $X$ satisfies the $\kappa$-Baire property.

Every subspace of $X$ satisfies the $\kappa$-Baire property.

If $f:\mathcal{T}\rightarrow M$ is a surjective frame homomorphism, then $M$ is $\kappa$-distributive.

If $f:\mathcal{T}\rightarrow B$ is a surjective frame homomorphism and $B$ is a complete Boolean algebra, then $B$ is
$\kappa$-distributive.